Underflow Exception when trying to bind small double to BINARY_DOUBLE

Question

I get exception when trying to insert small double value (1.0E-131) into BINARY_DOUBLE column via jdbc.
The exception I get:
Exception in thread "main" java.sql.SQLException: Underflow Exception trying to bind 1.0000000000000001E-131
at oracle.jdbc.driver.DoubleBinder.bind(OraclePreparedStatement.java:13856)
at oracle.jdbc.driver.OraclePreparedStatement.setupBindBuffers(OraclePreparedStatement.java:2866)
at oracle.jdbc.driver.OraclePreparedStatement.processCompletedBindRow(OraclePreparedStatement.java:2151)
at oracle.jdbc.driver.OraclePreparedStatement.executeInternal(OraclePreparedStatement.java:3280)
at oracle.jdbc.driver.OraclePreparedStatement.execute(OraclePreparedStatement.java:3390)
at dbAccess.main(dbAccess.java:18)
According to oracle documentation the smallest positive number the BINARY_DOUBLE column can store is 2.22507485850720E-308.
I'm using oracle 10g r2 and thin jdbc driver.
Any ideas?

Answer 1

That is expected since a double is converted to a Oracle NUMBER before passing it to the database and Oracle NUMBER does not support 1.0e-131, the range is 1E-130 .. 10E125.
Look at the SQL output below,
SQL> select to_char(to_number(1.0e-131)) from dual;
T
0
SQL> select to_char(to_number(1.0e-130)) from dual;
TO_CHAR(TO_NUMBER(1.0E-130))
1.000000000000000000000000000000000E-130
==
Use oracle.sql.BINARY_DOUBLE for numbers like this.
eg:
preparedStatement.setBINARY_DOUBLE(1, new oracle.sql.BINARY_DOUBLE(1.0e-131));
==
hope this helps.
Ashok